Solve the initial value problem displaystyle{leftlbracebegin{matrix}{y}text{}+{16}{y}= cos{{left({4}{t}right)}}{y}{left({0}right)}={1}{y}'{left({0}right)}={1}end{matrix}right.} using the Laplace transform.

Question
Laplace transform
asked 2020-10-23
Solve the initial value problem \(\displaystyle{\left\lbrace\begin{matrix}{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}}\\{y}{\left({0}\right)}={1}\\{y}'{\left({0}\right)}={1}\end{matrix}\right.}\) using the Laplace transform.

Answers (1)

2020-10-24
Step 1
Given: \(\displaystyle{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}\)
Step 2
Explanation:
\(\displaystyle{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}\)
Taking Laplace transform on both sides
\(\displaystyle{L}{\left[{y}{''}+{16}{y}\right]}={L}{\left[ \cos{{\left({4}{t}\right)}}\right]}\)
\(\displaystyle{L}{\left({y}\text{}{\left({t}\right)}\right)}+{L}{\left({16}{y}{\left({t}\right)}\right)}={L}{\left[ \cos{{\left({4}{t}\right)}}\right]}\ldots{\left({i}\right)}\)
We know that if \(L(y(t))=y(s)\)
then \(\displaystyle{L}{\left({y}{''}{\left({t}\right)}\right)}={s}^{2}{y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\)
\(\displaystyle{L}{\left[ \cos{{\left({a}{t}\right)}}\right]}=\frac{s}{{{s}^{2}+{a}^{2}}}\)
Using these in equation (i), we get
\(\displaystyle{s}^{2}{y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{16}{y}{\left({s}\right)}=\frac{s}{{{s}^{2}+{4}^{2}}}\)
Using the given initial conditions \(y(0)=1, y'(0)=1\) then
\(\displaystyle{\left({s}^{2}+{16}\right)}{y}{\left({s}\right)}-{s}{\left({1}\right)}-{1}=\frac{s}{{{s}^{2}+{4}^{2}}}\)
\(\displaystyle{\left({s}^{2}+{4}^{2}\right)}{y}{\left({s}\right)}={1}+{s}+\frac{s}{{{s}^{2}+{4}^{2}}}\)
\(\displaystyle{y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{\left({s}^{2}+{4}^{2}\right)}{\left({s}^{2}+{4}^{2}\right)}}}\)
\(\displaystyle{y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{\left({s}^{2}+{4}^{2}\right)}^{2}}}\)
Taking inverse Laplace transform on both sides
\(\displaystyle{L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{\left({s}^{2}+{4}^{2}\right)}^{2}}\right)}\)
\(\displaystyle{y}{\left({t}\right)}={L}^{ -{{1}}}{\left(\frac{1}{{{s}^{2}+{4}^{2}}}\right)}+{L}^{ -{{1}}}{\left(\frac{s}{{{s}^{2}+{4}^{2}}}\right)}+{L}^{ -{{1}}}{\left(\frac{s}{{\left({s}^{2}+{4}^{2}\right)}^{2}}\right)}\)
\(\displaystyle=\frac{1}{{4}} \sin{{\left({4}{t}\right)}}+ \cos{{\left({4}{t}\right)}}+\frac{1}{{{2}\cdot{4}}}{t} \sin{{\left({4}{t}\right)}}\)
\(\displaystyle=\frac{1}{{4}} \sin{{\left({4}{t}\right)}}+ \cos{{\left({4}{t}\right)}}+\frac{1}{{8}}{t} \sin{{\left({4}{t}\right)}}\)
0

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